仅此而已

Mark

``` const getStr=(str,splitNum=3,split=',')=>{ return str.split('').reverse().map((v,i)=>{ if(i&&i%splitNum==0){ return `${v}${split}` } return v }).reverse().join(''); } getStr('10000000000',3,','); ```

不会正则就用粗旷的方式吧 ``` '1000000000000'.split('').reverse().map((v,i)=>{ return i%3==2?`,${v}`:v }).reverse().join('') ```

通过count计数的形式，判断接口是否全部完成 ``` class MultiRequest { constructor(props) { this.max = props.max || 10; //数量 this.urls = props.urls || []; //接口 this.count = 0; //总共完成数量 this.result = []; //接口请求完成集合 this.thenCbs = [];...

@ > ```js > function convert（list）{ > const res = [] > const map = list。减少（（RES，v）=>（RES [ v。ID ] = V，RES），{}） > 为（const的 项目 的列表）{ > 如果（项目。parentId的 === 0）{ >...

``` const convert = (list) => { if (!Array.isArray(list)) { return [] } list.forEach((v) => { const index = list.findIndex((_l) => { return v.parentId == _l.id }); if (index ==...

> ```js > function convert(list) { > const res = [] > const map = list.reduce((res, v) => (res[v.id] = v, res), {}) > for (const item of list) {...

> > > ```js > > > function convert(list) { > > > const res = [] > > > const map = list.reduce((res, v) => (res[v.id] = v, res),...

> > > > > ```js > > > > > function convert(list) { > > > > > const res = [] > > > > > const map...

I use the regular ` ` ` var regex=/(?=(YYYY|YY|MM|DD|HH|mm|ss|ms))/; ` ` ` already can be normal use, I want to know the following regular ` ` ` var regex=/(?=(YYYY|YY|MM|DD|HH|mm|ss|ms))/;` `...